unsigned

Consider the following piece of C code: #include <stdint.h> uint32_t inc(uint16_t x) { return x+1; } When compiled with gcc-4.4.3 with flags -std=c99 -march=core2 -msse4.1 -O2 -pipe -Wall on a pure x86_64 system, it produces movzwl %di,%eax inc %eax retq Now, unsigned overflow is predicted in C. I do not know much about x86_64 assembly, but as far as I can see the 16bit argument register is being moved to a 32bit register, which is incremented and returned. My question is, what if x == UINT16_MAX. An overflow would occur and the standard dictates x+1==0, right? However, given %eax is a 32bit

I would like to convert a java.net.InetAddress and I fight with the signed / unsigned problems. Such a pain. I read convert from short to byte and viceversa in Java and Why byte b = (byte) 0xFF is equals to integer -1? And as a result came up with: final byte [] pumpeIPAddressRaw = java.net.InetAddress.getByName (pumpeIPAddressName).getAddress (); final long pumpeIPAddress = ((pumpeIPAddressRaw [0] & 0xFF) << (3*8)) + ((pumpeIPAddressRaw [1] & 0xFF) << (2*8)) + ((pumpeIPAddressRaw [2] & 0xFF) << (1*8)) + (pumpeIPAddressRaw [3] & 0xFF); android.util.Log.i ( Application.TAG, "LOG00120: Setzte

方括号中的属性可以在DDL语句中控制COLUMN的详细属性 一、整数型 1.INT[(width)][UNSIGNED][ZEROFILL] MySQL用4 bytes存储INT型数据，其值在-2,147,483,648到2,147,483,647之间，如果选择了UNSIGNED类型，那么值在0到4,294,967,295。INT和INTEGER可以互换。(width)指定了数字的位数，如果实际的值超出了这个位数，那么(width)会被忽略。如果是UNSIGEND，通过指定ZEROFILL会由0在左侧占位补足 2.BOOLEAN 也可以写为BOOL或BIT，需要1个byte空间，用来存储boolean value，false(zero)或者true(nonzero)，等价于TINYINT(1) 3.TINYINT[(width)][UNSIGNED][ZEROFILL] 需要一个byte空间，值的范围在-127到128或0到255 4.SMALLINT[(width)][UNSIGNED][ZEROFILL] 需要2bytes空间，值的范围在-32768到32767或者0到65535 5.MEDIUMINT[(width)][UNSIGNED][ZEROFILL] 需要3个byte，范围在-8,388608到8,388,607或者0到16,777,215 6.BIGINT[

在学习ufldl课程时需要用到MNIST数据集，主页在 这里 。但由于该数据集为IDX文件格式，是一种用来存储向量与多维度矩阵的文件格式，不能直接读取。 mnist的结构如下 TRAINING SET LABEL FILE (train-labels-idx1-ubyte): [offset] [type] [ value ] [description] 0000 32 bit integer 0x00000801 ( 2049 ) magic number (MSB first ) 0004 32 bit integer 60000 number of items 0008 unsigned byte ?? label 0009 unsigned byte ?? label ........ xxxx unsigned byte ?? label The labels values are 0 to 9. TRAINING SET IMAGE FILE (train-images-idx3-ubyte): [offset] [type] [ value ] [description] 0000 32 bit integer 0x00000803 ( 2051 ) magic number 0004 32 bit integer 60000 number of images

#include <iostream> #include <string> using namespace std; void H_getMD5(const unsigned long Message[16]); static unsigned long State[4]; /* MD5值是由4个32bit的数组成的，也就是4个4Byte长的无符号整型 */ static unsigned long Count[2]; /* MD5算法除了要对原字符串按bit填充1和0之外，还需要在末尾附上64bit 的值，值是用字符串的bit长度 mod 2^64，这两整型是8byte，共64bit */ static unsigned char buffer[64]; /* MD5按每 512bit 一组来处理，这用来存放每组的 512bit */ /* Left_Shift表示，对x进行循环左移 n 位 */ #define Left_Shift(x,n) (((x)<<(n))|((x)>>(32-(n)))) /* F、G、H、I 4个函数是MD5的4个基本的非线性函数 */ #define F(x,y,z) (((x)&(y))|((~x)&(z))) #define G(x,y,z) (((x)&(z))|((y)&(~z))) #define H(x,y,z) ((x)^(y)^

In C++ we can make primitives unsigned . But they are always positive. Is there also a way to make unsigned negative variables? I know the word unsigned means "without sign", so also not a minus (-) sign. But I think C++ must provide it. No. unsigned can only contain nonnegative numbers. If you need a type that only represent negative numbers, you need to write a class yourself, or just interpret the value as negative in your program. (But why do you need such a type?) unsigned integers are only positive. From 3.9.1 paragraph 3 of the 2003 C++ standard: The range of nonnegative values of a

In Java the int type is signed, but it has a method that compares two ints as if they were unsigned: public static int compareUnsigned(int x, int y) { return compare(x + MIN_VALUE, y + MIN_VALUE); } It adds Integer.MIN_VALUE to each argument, then calls the normal signed comparison method, which is: public static int compare(int x, int y) { return (x < y) ? -1 : ((x == y) ? 0 : 1); } How does adding MIN_VALUE to each argument magically make the comparison unsigned? This technique works for any size of integer, but I'll use an 8-bit byte-sized integer to explain, because the numbers are smaller

类型 范围 字节 占位符 char -128~127 1 %c　　%hhd unsigned char 0~255 1 %c　　%hhu short -32768~32767 2 %hd unsigned short 0~65535 2 %hu int -2^31~2^31-1 4 %d unsigned （int） 0~2^32-1 %u long （int） -2^31~2^31-1(x86)　　-2^63~2^63-1(x64) 4　　8 %ld unsigned long （int） 0~2^32-1　　　　　　　　0~2^64-1 4　　8 %lu long long （int） -2^63~2^63-1 8 %lld unsigned long long （int） 0~2^64-1 8 %llu float 4 %f　　%g double %lf　　%lg 来源： https://www.cnblogs.com/smilyD/p/11906131.html

设一个基准情况；再调用 第二个情况由第一个情况递推，再调用； 达到某个情况，返回值。 有5个人围坐在一起，问第5个人多大年纪，他说比第4个人大2岁；问第4个人，他说比第3个人大2岁；问第3个人，他说比第2个人大2岁；问第2个人，他说比第1个人大2岁。第1个人说自己10岁，请利用递归法编程计算并输出第5个人的年龄。 include <stdio.h> unsigned int ComputeAge(unsigned int n); main() { unsigned int n = 5; printf("The 5th person's age is %d\n", ComputeAge(n)); } /* 函数功能：用递归算法计算年龄 */ unsigned int ComputeAge(unsigned int n) { unsigned int age; if (n == 1) { age = 10; } else { age = ComputeAge(n - 1) + 2; } return age; } 来源： https://www.cnblogs.com/dosu/p/11905195.html